Binomial and Poisson Random Variables – Solutions STAT-UB.0103 – Statistics for Business Control and Regression Models
Binomial random variables
1. A certain coin has a 25% of landing heads, and a 75% chance of landing tails. (a) If you flip the coin 4 times, what is the chance of getting exactly 2 heads?
Solution: There are 6 outcomes whith exactly 2 heads:
HHT T, HT HT, HT T H, T HHT, T HT H, T T HH.
By independence, each of these outcomes has probability (.25)2(.75)2. Thus, P(exactly 2 heads out of 4 flips) = 6(.25)2(.75)2.
(b) If you flip the coin 10 times, what is the chance of getting exactly 2 heads?
Solution: Rather than list all outcomes, we will use a counting rule. There are 102 ways of choosing the positions for the two heads; each of these outcomes has probability (.25)2(.75)8. Thus,
P(exactly 2 heads out of 10 flips) =10 2
(.25)2(.75)8.
2. Suppose that you are rolling a die eight times. Find the probability that the face with two spots comes up exactly twice.
Solution: Let X be the number of times that we get the face with two spots. This is a binomial random variable with n = 8 and p = 16. We compute
P (X = 2) =n 2 p2(1 − p)n−2 =8 2 1 6 2 5 6 6 ≈ 0.26.
3. The probability is 0.04 that a person reached on a “cold call” by a telemarketer will make a purchase. If the telemarketer calls 40 people, what is the probability that at least one sale with result?
Solution: Let X be the number of sales. This is a binomial random variable with n = 40 and p = 0.04. Thus, P (X ≥ 1) = 1 − P (X < 1) = 1 − P (X = 0) = 1 −n 0 p0(1 − p)n−0 = 1 − (0.96)40 ≈ .805
4. A new restaurant opening in Greenwich village has a 30% chance of survival during their first year. If 16 restaurants open this year, find the probability that
(a) exactly 3 restaurants survive.
Solution: Let X be the number that survive. This is a binomial random variable with n = 16 and p = 0.3. Therefore,
P (X = 3) =16 3
(0.3)3(1 − 0.3)(16 − 3) = .146
(b) fewer than 3 restaurants survive. Solution: P (X < 3) =16 0 (0.3)0(0.7)16+16 1 (0.3)1(0.7)15+16 2 (0.3)2(0.7)14 = .099.
(c) more than 3 restaurants survive. Solution:
P (X > 3) = 1 − P (X ≤ 3) = 1 − (.099 + .146) = .754.
Poisson random variables
5. The number of calls arriving at the Swampside Police Station follows a Poisson distribution with rate 4.6/hour. What is the probability that exactly six calls will come between 8:00 p.m. and 9:00 p.m.?
Solution: Let X be the number of calls that arrive between 8:00 p.m. and 9:00 p.m. This is a Poisson random variable with mean
λ = E(X) = (4.6 calls/hour)(1 hour) = 4.6 calls. Thus, P (X = 6) = λ 6 6!e −λ = (4.6) 6 6! e −4.6 .
6. In the station from Problem 5, find the probability that exactly 7 calls will come between 9:00 p.m. and 10:30 p.m.
Solution: Let X be the number of calls that arrive between 9:00 p.m. and 10:30 p.m. This is a Poisson random variable with mean
λ = E(X) = (4.6 calls/hour)(1.5 hours) = 6.9 calls. Thus, P (X = 7) = λ 7 7!e −λ = (6.9) 7 7! e −6.9 .
7. Car accidents occur at a particular intersection in the city at a rate of about 2/year. Estimate the probability of no accidents occurring in a 6-month period.
Solution: Let X be the number of car accidents. This is Poisson random variable with mean
8. In the intersection from Problem 7, estimate the probability of two or more accidents occurring in a year.
Solution: Let X be the number of car accidents. This is Poisson random variable with mean
λ = E(X) = (2 accidents/year)(1.0 years) = 2 accident. Thus, P (X ≥ 2) = 1 − P (X < 2) = 1 − λ 0 0!e −λ + λ 1 1!e −λ ≈ .594.
Empirical rule with Binomial and Poisson random variables
9. If X is a Poisson random variable with λ = 225, would it be unusual to get a value of X which is less than 190?
Solution: Set
µ = E(X) = λ = 255, σ = sd(X) =√λ = 15.
Define z to be the number of standard deviations above the mean that 190 is, i.e. 190 = µ + σz. Then, z = 190 − µ σ = −35 15 ≈ −2.33.
A value of X which is below 190 is more than 2.33 standard deviations below the mean of X. The empirical rule tells us that observations more than 2 standard deviations away from the mean are unusual (they occur less than 95% of the time). Therefore, values of X below 190 are unusual.
10. The probability is 0.10 that a person reached on a “cold call” by a telemarketer will make a purchase. If the telemarketer calls 200 people, would it be unusual for them to get 30 purchases?
Solution: Let X be the number of purchases. This is a Binomial random variable with size n = 200 and success probability p = 0.10. Thus, the expectation and standard deviation of X are
µ = np = (200)(.10) = 20
σ =pnp(1 − p) =p(200)(.10)(.90) =√18 ≈ 4.2
Since np ≥ 15 and np(1 − p) ≥ 15, the distribution of X can be approximated by the empirical rule. Using the empirical rule approximation, 95% of the time, X will be in the
